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\def\lf{\ \hfil\break}       % Neue Zeile ohne Einr"ucken, 'linefeed'
\def\cl{\centerline}
\def\LF{\medskip\noindent}   % Neue Zeile mit breiterem Zwischenraum


\nopagenumbers




\cl{\bf Boy's Surface (Bryant-Kusner)}

\lf
See M\"obius Strip first. The non-orientable surfaces are {\it one-sided} 
and this property can best be understood, if one starts from a M\"obius Strip. 

\noindent
The Klein Bottle is easier to visualize
than the Boy's surface. Each meridian of the Boy's surface is the
center line of a narrow M\"obius band. To see this, select  
``Set u,v Ranges'' from the Settings menu and for example change 
umin to $ - 0.998$, and  vmin to $6.1$.

\noindent
The ``equator'' of the Boy's surface is a different kind of M\"obius band,
it has {\it three} half-twists instead of one. The standard morph
begins with this M\"obius band and widens it until the Boy's
surface is complete: 

$ aa = 0.5$, vmin $= 0$, vmax $= 2\pi$, umax $= 1$, and 

$0.9 \ge \hbox{umin} \ge 0.002$

\noindent
``Boy's surface'' is really a family of surfaces. Boy, in his dissertation
under Hilbert, constructed this surface
as the first known {\it immersion} of the projective plane.
Being nonorientable implies that no embedding is possible. Boy's surface
has, besides its self-intersection curves, only one more singularity,
namely a triple point. Boy's construction was topological.
Ap\'ery has found algebraically embedded ``Boy's surfaces''. These carry
one-parameter families of ellipses.

\noindent
The Bryant-Kusner Boy's surfaces are obtained by an inversion from
a minimal surface in $\Bbb R^3$. The minimal surface is an immersion
of $\Bbb S^2 - \{6\ \hbox{points}\}$ such that antipodal points have the same
image in $\Bbb R^3$. The six punctures are three antipodal pairs, and
the minimal surface has so called {\it planar ends} at these punctures.
In this context it is important that the inversion of a planar end has
the puncture that can be closed {\it smoothly} by adding one point.
The closing of the three pairs of antipodal ends thus gives a triple
point on the surface obtained by inversion.  Explicitly let 

\noindent
  $M(z) = \Re( a(z) V(z) ) + (0,0,1/2)$, where 

\noindent
$a(z) = \left(z^3 - z^{-3} + \sqrt{5} \right) ^{-1}$ and 

\noindent
$V(z) = \left( i ( z^2 - z^{-2} ) , z^2 + z^{-2},{2 i\over 3} ( z^3 + z^{-3} ) \right)$.

\noindent
Then $Boys(z)$ is obtained by inverting $M(z)$ in the unit sphere:
$$Boys(z) := {M(z)\over || M(z) ||^2}.$$
\noindent H.K.





\bye
